$\lim_{x\to 2}\dfrac{x^3-2x^2}{x^2-4}=$
Answer: Substituting $x=2$ into $\dfrac{x^3-2x^2}{x^2-4}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression on our hands, let's try to simplify it. $\dfrac{x^3-2x^2}{x^2-4}$ can be simplified as $\dfrac{x^2}{x+2}$, for $x\neq 2$. This means that the two expressions have the same value for all $x$ -values (in their domains) except for $2$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{x^3-2x^2}{x^2-4}=\dfrac{x^2}{x+2}$ for all $x$ -values in the interval $(1,3)$ except for $x=2$. Therefore, $\lim_{x\to 2}\dfrac{x^3-2x^2}{x^2-4}=\lim_{x\to 2}\dfrac{x^2}{x+2}=1$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to 2}\dfrac{x^3-2x^2}{x^2-4}=1$.